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Topic: Square Polynomial (Read 795 times) |
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ThudnBlunder
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Square Polynomial
« on: Jan 3rd, 2007, 9:44pm » |
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Find integers a,b,c such that P(x) = ax2 + bx + c is not the square of a linear polynomial but P(n) is a perfect square for n = 1 to 4.
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Miles
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Cambridge, England
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Re: Square Polynomial
« Reply #1 on: Jan 4th, 2007, 6:58am » |
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Could have a = -36, b = 180, c = -95, which gives P(1) = 49, P(2) = 121, P(3) = 121, P(4) = 49, and it is clear that P is not a square of a linear term as a is negative.
I got this when I thought about plotting y = P(x) and realised that any quadratic going through the points
(1,m^2), (2,n^2), (3,n^2) and (4,m^2) where n > m >=0 would do the trick. This gives a general result (but I'm not saying it covers all possible solutions) of
P(x) = -(1/2) * [(n^2-m^2)*x^2 - 5*(n^2-m^2)*x + 4n^2 - 6m^2]
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towr
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Re: Square Polynomial
« Reply #2 on: Jan 4th, 2007, 7:20am » |
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Here's a couple..
-8, 40, -32
-8, 40, -23
-6, 30, -20
-4, 20, -15
-2, 10, -8
2, -10, 12
6, -30, 40
8, -40, 48
[edit]
searching a bit further; the first one I found for which P(1) =/= P(4) and P(2) =/= P(3) :
a=-20, b=60, c=81 -> P([1,2,3,4]) = [121, 121, 81, 1]
[/edit]
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| « Last Edit: Jan 4th, 2007, 7:28am by towr » |
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pex
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Re: Square Polynomial
« Reply #3 on: Jan 4th, 2007, 7:21am » |
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There are many more solutions. I found one (by programming, as usual; never mind that this is the Putnam thread) that gives distinct squares for n = 1 to 7:
P(x) = -4980x2 + 37620x - 16511
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Miles
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Re: Square Polynomial
« Reply #4 on: Jan 4th, 2007, 8:10am » |
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Nice, but I do mind that this is the "Putnam" thread. Can anyone provide an algebraic method to derive these polynomials?
Is there a limit to k such that there exists P with P(1), P(2), ... P(k) all square?
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Barukh
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Re: Square Polynomial
« Reply #5 on: Jan 4th, 2007, 9:42am » |
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Let me present here one approach that may be considered “algebraic”.
Relax one condition of the problem (say x = 1). Then, three coefficients a, b, c are determined uniquely by three chosen numbers P(2) = p2, P(3) = q2, P(4) = r2, by solving a system of linear equations with 3 unknowns. The general solution, of course, will express the coefficients a, b, c in terms of p2, q2, r2 (I don’t present them here, but they are pretty easy to derive, even online).
Now bringing back the forth condition, P(1) = a + b + c = s2. Using the above mentioned expressions, we get the following equality:
3(p2 - q2) + r2 = s2
The easiest way is to set p = q, and then r = s. Letting p = 3, r = 1, we get (a,b,c) = (-4, 20, -15).
But we want something more interesting! One way to proceed is to choose p2 - q2 to be an odd number, and then find r, s by factoring the left-hand side. The problem with this is that you too often end up with the square of the linear polynomial! Nevertheless, without using any computing power, I’ve found the following triple: (p, q, r) = (4, 1, 22), which produces the polynomial (249, -1260, 1540) with 4 disitinct squares 529, 16, 1, 484.
Unfortunately, this method doesn’t generalize to greater number of points.
h
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| « Last Edit: Jan 4th, 2007, 9:47am by Barukh » |
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ThudnBlunder
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Re: Square Polynomial
« Reply #6 on: Jan 4th, 2007, 10:18am » |
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on Jan 4th, 2007, 8:10am, Miles wrote:Nice, but I do mind that this is the "Putnam" thread. Can anyone provide an algebraic method to derive these polynomials?
Is there a limit to k such that there exists P with P(1), P(2), ... P(k) all square? |
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I also sorta mind.
Let P(x) = d(2x - 5)2 + e
Then
P(1) = P(4) = 9d + e = p2
P(2) = P(3) = d + e = q2
Hence
8d = p2 - q2
8e = 9q2 - p2
eg, putting p = 1, q = 3 gives d = -1, e = 10
and P(x) = -4x2 + 20x - 15
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| « Last Edit: Jan 5th, 2007, 7:03am by ThudnBlunder » |
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ThudnBlunder
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Re: Square Polynomial
« Reply #7 on: Jan 4th, 2007, 10:39am » |
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For 8 squares we can try 2x-9 instead of 2x-5
We then get
P(1) = P(8) = 49d + e = r2
P(2) = P(7) = 25d + e = s2
P(3) = P(6) = 9d + e = p2
P(4) = P(5) = d + e = q2
Giving
24d = 3p2 - 3q2 = r2 - s2
24e = 27q2 - 3p2 = 49s2 - 25r2
And only now should we unleash Pavlov's dogs, the CS nerds. ;)
Ha! Too slow.
d = -105
e = 5434
p = 67
q = 73
r = 17
s = 53
and
P(x) = -420x2 + 3780x - 3071
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| « Last Edit: Jan 5th, 2007, 7:03am by ThudnBlunder » |
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towr
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Re: Square Polynomial
« Reply #8 on: Jan 4th, 2007, 11:20am » |
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If we set the first two squares to hit at 0 and 1, we get the system
a + b + c = 0
4a + 2b + c = 1
9a + 3b + c = n2
16a + 4b + c = m2
from which we can derive
a = (n2 - 2)/2
b = (8 - 3n2)/2
c = n2 - 3
n2 = (m2 + 3)/3
from which we can see
n is even
and because we dont' want a square of a linear polynomial (and thus 4ac =/=b2)
n2 (n2 - 8) =/= -16 -> n=/= +-2
Now for
n2 = (m2 + 3)/3
We can find a recurrence
N1 = 1
N2=2
Nk = 4Nk-1 - Nk-2
And so we get solutions for
n m
26 45
362 627
5042 5733
70226 121635
etc
The first gives
a=337
b=-1010
c=673
NB, the first with 8 squares:
a=-420, b=3780, c=-3071
(for a,b,c under 5000 there's no longer sequence)
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| « Last Edit: Jan 4th, 2007, 11:23am by towr » |
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ThudnBlunder
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Re: Square Polynomial
« Reply #10 on: Jan 4th, 2007, 11:33am » |
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on Jan 4th, 2007, 11:26am, towr wrote:
I already had it several hours ago.
But you seemed mindfull against such results..
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The margin doesn't count. 
OK, I believe you.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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ThudnBlunder
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Re: Square Polynomial
« Reply #11 on: Jan 5th, 2007, 9:08am » |
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And
d = 570570
e = 4406791
p = 3089
q = 2231
r = 5689
s = 4321
gives
2282280x2 - 20540520x + 50622961
P(1) = P(8) = 56892
P(2) = P(7) = 43212
P(3) = P(6) = 30892
P(4) = P(5) = 22312
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